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Let f(x) = | x | be the absolute value function.
1. Why is the range of this function {y | y >= 0}?
2. Why does this function have an absolute minimum of 0 at x = 0?
Can the answer be that (-infinity, 0) and (0, infinity) meet at the origin where x = 0?
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Have you tried looking at the graph?
For positive x the graph is the same as y = x.
For negative x the graph is the same as y = -x
So, putting these together, you get a V shaped graph with the origin at the bottom of the V.
Your last line suggests you're looking at a different graph. ( y = 1/x maybe)
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Have you tried looking at the graph?
For positive x the graph is the same as y = x.
For negative x the graph is the same as y = -x
So, putting these together, you get a V shaped graph with the origin at the bottom of the V.
Your last line suggests you're looking at a different graph. ( y = 1/x maybe)
Bob
I was looking at the V-shape graph before posting this question (unless my eyes wondered away from the absolute value function to the reciprocal function). In any case, I need extra practice finding the domain range of graphs.
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Extra practice.
(1) y = x^2
(2) y = x^3 - 2x
(3) y = x^2/(x+4)
(4) y = sqrt(100-x^2)
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Extra practice.
(1) y = x^2
(2) y = x^3 - 2x
(3) y = x^2/(x+4)
(4) y = sqrt(100-x^2)
Bob
Ok. I will work on these and show my effort here.
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Extra practice.
(1) y = x^2
(2) y = x^3 - 2x
(3) y = x^2/(x+4)
(4) y = sqrt(100-x^2)
Bob
For y = x^2
Vertex = (0, 0) = absolute minimum
Range = all non-negative real numbers
Domain = all real numbers
No absolute maximum.
For y = x^3 - 2x
Range = Domain = all real numbers.
No absolute minimum.
No absolute maximum.
For y = (x^2)/(x + 4)
Domain = all real numbers except x cannot be -4.
This function is not one to one. So, it does not have an inverse.
I say no range.
I an not sure about absolute minimum and absolute maximum values.
For y = sqrt{100 - x^2}
Domain: -10 <= x <= 10
Range: 0 <= y <= 10
Absolute maximum at the point (0, 10).
No absolute minimum.
You say?
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Excellent attempt. 3/4 perfect. Number 3 is very tricky. Without a graph I'd have difficulty. It's the range that you need to work on. I recommend using the grapher.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Excellent attempt. 3/4 perfect. Number 3 is very tricky. Without a graph I'd have difficulty. It's the range that you need to work on. I recommend using the grapher.
Bob
I will try 3 again.
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